I recently learned of the result by Carleson and Hunt (1968) which states that if $f \in L^p$ for $p > 1$, then the Fourier series of $f$ converges to $f$ pointwise-a.e. Also, Wikipedia informs me that if $f \in L^p$ for $1 < p < \infty$, then the Fourier series of $f$ converges to $f$ in $L^p$. Either of these results implies that if $f \in L^p$ for $1 < p < \infty$, then the Fourier series of $f$ converges to $f$ in measure.

My first question is about the $p = 1$ case. That is:

If $f \in L^1$, will the Fourier series of $f$ converge to $f$ in measure?

I also recently learned that there exist functions $f \in L^1$ whose Fourier series diverge (pointwise) everywhere. Moreover, such a Fourier series may converge (Galstyan 1985) or diverge (Kolmogorov?) in the $L^1$ metric.

(Notes: Here, I mean the Fourier series with respect to the standard trigonometric system. I am also referring only to the Lebesgue measure on [0,1]. Of course, if anyone knows any more general results, that would be great, too.)

1 Answer
1

The answer to your first question is no. There is an L^1 function with Fourier series not converging in measure.

In the Kolmogorov example of an L^1 function $f$ with a.e. divergent Fourier series, there is in fact a set of positive measure E and a subsequence n_k such that for all x in E, the absolute values of the partial sums S_{n_k} of the Fourier series goes to infinity with k.

$$\forall x\in E,\ \ |S_{n_k}f(x)|\rightarrow \infty$$

This can be checked from the construction of f in the original article of Kolmogorov, in its selected works.

If $S_nf$ converges in measure, then $S_{n_k}f$ must also converges in measure. This implies that there is a subsequence $n_{k_l}$ such that $S_{n_{k_l}}f(x)$ converges a.e. x, a contradiction.